Chromatography Forum

Lets say one uses 0.015 N H2SO4 as eluant in mobile phase for cation exchange chromatography. If my rough calculation is correct (and it might not be, correct me if I'm wrong), pH should be around 1.8 to 2. Then suppose one wants to determine Na+ at a ppb level, but the injection solvent has a pH < 1, or potentially 10X or thereabouts more concentrated in hydronium ion than the eluant. Is that in itself a cause for chromatographic problems in finding Na+ at trace levels? It seems that the amount of Na+ in the diluent would be so overwhelmed by the amount of H+ in the diluent, that problems would ensue in detection of Na+. Again, my calculation might be wrong, but I think we'd be dealing with ~ 50 ppm of H+ in an environment where we were looking for Na+ at a ppb level. The Na+ would be "lost in the crowd" of H+ and how would we ever find it? Bottom line - if the pH of the injection media is a good bit lower than that of the eluant, will that cause problems in looking for Na+?

Bottom line - if the pH of the injection media is a good bit lower than that of the eluant, will that cause problems in looking for Na+?

It can, but not for the reason you suggest.

Any analytical HPLC separation with a linear response is done under approximately "infinite dilution" conditions. The hydronium concentration in your mobile phase is already about 11 orders of magnitude higher than the Na+ you want to detect*. One order of magnitude more or less isn't really going to matter a whole lot.

The potential problem is more with the fact that your diluent is stronger than your mobile phase. This has the potential to cause peak shape problems because the analyte will not equilibrate properly with the mobile phase at the head of the column (the same thing can happen in revesed-phase if the diluent is higher in organic solvent than the mobile phase). If your sample volume is small, this may not be an issue, but the usual way to get super-low detection is to inject a large sample.

*Someone check my math:
1 ppb ≈ 10^-9 g/mL (assuming water at 1 g = 1 mL)
≈ 10^-12g/L
≈ (10^-12)/23 moles/L
≈ 0.5 X 10^-12 M
≈ 5 X 10^-13 N (Na+ is monovalent)
divide by 1.5 X 10^-2 N acid ≈ 3 X 10^-11